Trig question about equation of a circle..

the circle is centered at (7, -2) and passes through (-10,0) i know that the equation for the circle can be found using the first points= (x-7)^2+(y+2)^2=r^2....but is r=-10? or what am i supposed to do with the (-10,0) to find r?

Okay.

So you got the first half of the equation right. But the radius is completely off. The circle isn't centered at (0,0), so you can't just get the radius like that.

You need to find the distance between the center (7,-2) and the point (-10,0).

You can find that by using the distance formula:
sqrrt([first x coord - second x coord]^2 + [first y coord - second y coord]^2)

You can subtract the second from the first, doesn't matter. As long as you have them in the same order for both x and y.

If we plug the values into the equation, we get:

sqrrt([7 - (-10)]^2 + [-2 - 0]^2)
sqrrt(17^2 + (-2)^2)
sqrrt(289 + 4)
sqrrt(293) = radius

We don't need to find the actually value since you'll just square it in the final equation, bringing it to a 293.

The FINAL EQUATION:

(x - 7)^2 + (y + 2)^2 = 293

Recall that the distance between the center of the circle and any point on the circle is the radius of circle, so to find r, you can use the distance formula to find the distance between (7, -2) and (-10, 0). Alternatively, you can use the fact that since (-10, 0) lies on the circle, x = -10 and y = 0 satisfy the equation of the circle. Therefore, plugging in x = -10 and y = 0 yields:
r^2 = (-10 - 7)^2 + (0 + 2)^2 = 17^2 + 2^2 = 293,

giving the equation of the circle to be:
(x - 7)^2 + (y + 2)^2 = 293.

I hope this helps!

(-10,0) is a point on the circle.
(7, -2) is the center of the circle.
Therefore, the distance between these points is the radius of the circle.

Use the distance formula to determine r.

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use the distance formula to find the distance from the center to a point on the circle.
That distance is the definition of the radius.